Javascript 合并2个对象数组
让我们看一个例子Javascript 合并2个对象数组,javascript,jquery,arrays,Javascript,Jquery,Arrays,让我们看一个例子 var arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"}); var arr2 = new Array({name : "childs", value: '5'}, {name: "
var arr1 = new Array({name: "lang", value: "English"},
{name: "age", value: "18"});
var arr2 = new Array({name : "childs", value: '5'},
{name: "lang", value: "German"});
我需要合并这两个对象数组并创建以下数组:
var arr3 = new Array({name: "lang", value: "German"},
{name: "age", value: "18"},
{name : "childs", value: '5'});
是否有任何JavaScript或jQuery函数可以执行此操作
$.extend
不适合我。它回来了
var arr4 = new Array({name : "childs", value: '5'},
{name: "lang", value: "German"});
jQuery合并怎么样
JSFIDLE示例:您可以使用对象收集属性,同时替换重复项,然后将该对象展开/展平回数组。大概是这样的:
function merge(args) {
args = Array.prototype.slice.call(arguments);
var o = { };
for(var i = 0; i < args.length; ++i)
for(var j = 0; j < args[i].length; ++j)
o[args[i][j].name] = args[i][j].value;
return o;
}
function expand(o) {
var a = [ ];
for(var p in o)
if(o.hasOwnProperty(p))
a.push({ name: p, value: o[p]});
return a;
}
var arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"});
var arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});
var arr3 = expand(merge(arr1, arr2));
在a
中提供与arr3
中相同的内容,其中“德语”替换为“煎饼”
这种方法确实假设您的对象都具有相同的{name:…,value:…}
形式
您可以看到它在这里工作(请打开控制台):合并两个阵列:
var arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"});
var arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});
var result=arr1.concat(arr2);
// result: [{name: "lang", value: "English"}, {name: "age", value: "18"}, {name : "childs", value: '5'}, {name: "lang", value: "German"}]
正在合并两个数组,但“name”的值不重复:
var arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"});
var arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});
var i,p,obj={},result=[];
for(i=0;i<arr1.length;i++)obj[arr1[i].name]=arr1[i].value;
for(i=0;i<arr2.length;i++)obj[arr2[i].name]=arr2[i].value;
for(p in obj)if(obj.hasOwnProperty(p))result.push({name:p,value:obj[p]});
// result: [{name: "lang", value: "German"}, {name: "age", value: "18"}, {name : "childs", value: '5'}]
var arr1=新数组({name:“lang”,value:“English”},{name:“age”,value:“18”});
var arr2=新数组({name:'childs',value:'5'},{name:'lang',value:'German'});
变量i,p,obj={},结果=[];
对于(i=0;i更新日期为2019年10月12日)
新版本仅基于较新的Javascript,不需要任何第三方库。
constmergebyproperty=(目标、源、属性)=>{
source.forEach(sourceElement=>{
让targetElement=target.find(targetElement=>{
返回sourceElement[prop]==targetElement[prop];
})
targetElement?Object.assign(targetElement,sourceElement):target.push(sourceElement);
})
}
var target/*arr1*/=[{name:“lang”,value:“English”},{name:“age”,value:“18”}];
var source/*arr2*/=[{name:“childs”,value:'5'},{name:“lang”,value:“德语”}];
mergeByProperty(目标、源、“名称”);
console.log(target)
我也遇到了同样的问题,基于此,我扩展了下划线库,并添加了我需要的更多功能。下面是示例
希望它有用!
问候!另一个版本使用:
var arr1=新数组({name:“lang”,value:“English”},{name:“age”,value:“18”});
var arr2=新数组({name:'childs',value:'5'},{name:'lang',value:'German'});
var arr=arr1.concat(arr2).reduce(函数(prev,current,index,array){
如果(!(上一个键中的current.name)){
上一个键[当前名称]=索引;
上一结果推送(当前);
}
否则{
上一个结果[上一个关键字[当前名称]]=当前;
}
返回上一个;
},{result:[],键:{});
document.getElementById(“output”).innerHTML=JSON.stringify(arr,null,2);
var newArray=yourArray.concat(otherArray);
console.log('concatated newArray:',newArray);
如果您想在JavaScript中合并两个对象数组。您可以使用这一行技巧
Array.prototype.push.apply(arr1,arr2);
比如说
var arr1 = [{name: "lang", value: "English"},{name: "age", value: "18"}];
var arr2 = [{name : "childs", value: '5'}, {name: "lang", value: "German"}];
Array.prototype.push.apply(arr1,arr2);
console.log(arr1); // final merged result will be in arr1
输出:
[{"name":"lang","value":"English"},
{"name":"age","value":"18"},
{"name":"childs","value":"5"},
{"name":"lang","value":"German"}]
arr3 = [
{"name":"lang","value":"German"},
{"name":"age","value":"18"},
{"name":"childs","value":"5"},
{"name":"lang","value":"German"}
]
我最近被这个问题难住了,我来到这里是希望得到一个答案,但被接受的答案使用了2 for in循环,这是我不喜欢的。我最终成功地制作了自己的。不依赖任何库:
function find(objArr, keyToFind){
var foundPos = objArr.map(function(ob){
return ob.type;
}).indexOf(keyToFind);
return foundPos;
}
function update(arr1,arr2){
for(var i = 0, len = arr2.length, current; i< len; i++){
var pos = find(arr1, arr2[i].name);
current = arr2[i];
if(pos !== -1) for(var key in arr2) arr1[pos][key] = arr2[key];
else arr1[arr1.length] = current;
}
}
函数查找(objArr,keyToFind){
var foundPos=objArr.map(函数(ob)){
返回ob.type;
}).indexOf(keyToFind);
返回foundPos;
}
功能更新(arr1、arr2){
对于(var i=0,len=arr2.length,current;i
这也保持了arr1的顺序。与lodash:
_.uniqBy([...arr1, ...arr2], 'name')
根据@YOU的回答,但遵守秩序:
var arr3 = [];
for(var i in arr1){
var shared = false;
for (var j in arr2)
if (arr2[j].name == arr1[i].name) {
arr3.push(arr1[j]
shared = true;
break;
}
if(!shared) arr3.push(arr1[i])
}
for(var j in arr2){
var shared = false;
for (var i in arr1)
if (arr2[j].name == arr1[i].name) {
shared = true;
break;
}
if(!shared) arr3.push(arr2[j])
}
arr3
我知道这种解决方案效率较低,但如果您想保持秩序,同时仍要更新对象,就必须这样做。对于那些正在尝试现代事物的人:
var奇数=[{
名称:“1”,
arr:“奇数”
},
{
名称:“3”,
arr:“奇数”
}
];
var偶数=[{
名称:“1”,
arr:“偶数”
},
{
名称:“2”,
arr:“偶数”
},
{
名称:“4”,
arr:“偶数”
}
];
// ----
//ES5使用Array.filter和Array.find
函数合并(a、b、prop){
var reduced=a.filter(函数(aitem){
return!b.find(函数(bitem){
返回aitem[prop]==bitem[prop];
});
});
收益减少。concat(b);
}
log(“ES5”,merge(奇数、偶数,“name”);
// ----
//ES6箭头函数
函数合并(a、b、prop){
var reduced=a.filter(aitem=>!b.find(bitem=>aitem[prop]==bitem[prop]))
收益减少。concat(b);
}
log(“ES6”,merge(奇数、偶数,“name”);
// ----
//ES6单班轮
var merge=(a,b,p)=>a.filter(aa=>!b.find(bb=>aa[p]==bb[p])).concat(b);
log(“ES6一行程序”,合并(奇数、偶数,“名称”);
//结果
//(b数组中的内容替换a数组中的内容)
// [
// {
//“名称”:“3”,
//“arr”:“奇数”
// },
// {
//“名称”:“1”,
//“arr”:“偶数”
// },
// {
//“名称”:“2”,
//“arr”:“偶数”
// },
// {
//“名称”:“4”,
//“arr”:“偶数”
// }
//]
我编写了一个jQuery插件,用一个键合并两个对象数组。请记住,这会修改目标数组
(函数($){
$.extendObjectArray=函数(destArray、srcArray、key){
对于(var index=0;index0){
var existingIndex=destArray.indexOf(existObject[0]);
$.extend(true,destArray[existingIndex],srcObject);
}否则{
destArray.push(srcObject);
}
}
返回数组;
};
})(jQuery);
变量arr1=[
{name:“lang”,value:“English”},
{名称:
_.uniqBy([...arr1, ...arr2], 'name')
var arr3 = [];
for(var i in arr1){
var shared = false;
for (var j in arr2)
if (arr2[j].name == arr1[i].name) {
arr3.push(arr1[j]
shared = true;
break;
}
if(!shared) arr3.push(arr1[i])
}
for(var j in arr2){
var shared = false;
for (var i in arr1)
if (arr2[j].name == arr1[i].name) {
shared = true;
break;
}
if(!shared) arr3.push(arr2[j])
}
arr3
var arr3 = _.uniqBy(arr1.concat(arr2), 'name'); // es5
let arr3 = _.uniqBy([...arr1, ...arr2], 'name'); // es6
const a = [{a: 1}, {b: 2}]
const b = [{a: 1}]
const result = a.concat(b) // [{a: 1}, {b: 2}, {a: 1}]
function merge(array1, array2, prop) {
return array2.map(function (item2) {
var item1 = array1.find(function (item1) {
return item1[prop] === item2[prop];
});
return Object.assign({}, item1, item2);
});
}
function mergeArrays(arrays, prop) {
const merged = {};
arrays.forEach(arr => {
arr.forEach(item => {
merged[item[prop]] = Object.assign({}, merged[item[prop]], item);
});
});
return Object.values(merged);
}
var arr1 = [
{ name: 'Bob', age: 11 },
{ name: 'Ben', age: 12 },
{ name: 'Bill', age: 13 },
];
var arr2 = [
{ name: 'Bob', age: 22 },
{ name: 'Fred', age: 24 },
{ name: 'Jack', age: 25 },
{ name: 'Ben' },
];
console.log(mergeArrays([arr1, arr2], 'name'));
arr1.filter(item => {
if (!arr2.some(item1=>item.name==item1.name)) {
return item
}
}).concat(arr2)
const extend = function*(ls,xs){
yield* ls;
yield* xs;
}
console.log( [...extend([1,2,3],[4,5,6])] );
merge(a, b, key) {
let merged = [];
a.forEach(aitem => {
let found = b.find( bitem => aitem[key] === bitem[key]);
merged.push(found? found: aitem);
});
return merged;
}
const merge = (a, b, key = "id") =>
a.filter(elem => !b.find(subElem => subElem[key] === elem[key]))
.concat(b);
merge(arr1, arr2, 'name');
var tx = [{"id":1},{"id":2}];
var tx1 = [{"id":3},{"id":4}];
var txHistory = tx.concat(tx1)
console.log(txHistory);
// output
// [{"id":1},{"id":2},{"id":3},{"id":4}];
var arr1 = new Array({name: "lang", value: "German"}, {name: "age", value: "18"});
var arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});
var arr3 = [...arr1, ...arr2];
arr3 = [
{"name":"lang","value":"German"},
{"name":"age","value":"18"},
{"name":"childs","value":"5"},
{"name":"lang","value":"German"}
]
function mergeObjectArraysRemovingDuplicates(firstObjectArray, secondObjectArray) {
return firstObjectArray.concat(
secondObjectArray.filter((object) => !firstObjectArray.map((x) => x.uniqueId).includes(object.uniqueId)),
);
}
const merge = (arr1, arr2, prop) => {
const resultMap = new Map(arr1.map((item) => [item[prop], item]));
arr2.forEach((item) => {
const mapItem = resultMap.get(item[prop]);
if (mapItem) Object.assign(mapItem, item);
else resultMap.set(item[prop], item);
});
return [...resultMap.values()];
};
const arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"});
const arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});
console.log(merge(arr1, arr2, "name"));
let mergeArray = arrA.filter(aItem => !arrB.find(bItem => aItem.name === bItem.name))